(\partial)/(\partial x)(ln(1-x-y)-x^{-1}-y^{-1})

解答

\frac{\partial}{\partial x} (ln (1 - x - y) - x^{- 1} - y^{- 1})

\frac{1}{x ^{2}} - \frac{1}{1 - x - y}

求解步骤

\frac{\partial}{\partial x} (ln (1 - x - y) - x^{- 1} - y^{- 1})

将

y

视为常数

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (ln (1 - x - y)) - \frac{\partial}{\partial x} (x^{- 1}) - \frac{\partial}{\partial x} (y^{- 1})

\frac{\partial}{\partial x} (ln (1 - x - y)) = - \frac{1}{1 - x - y}

\frac{\partial}{\partial x} (x^{- 1}) = - \frac{1}{x ^{2}}

\frac{\partial}{\partial x} (y^{- 1}) = 0

= - \frac{1}{1 - x - y} - (- \frac{1}{x ^{2}}) - 0

化简

= \frac{1}{x ^{2}} - \frac{1}{1 - x - y}

d er i v a t i v e f (x) = e^{- 5 x}

d er i v a t i v e y = x + 10

in v erse l a pl a ce \frac{( 5 )}{s ^{2} - 9}

l a pl a ce t r an s f or m t^{3} + 3 t^{2} + 3 t + 1

\frac{d}{d t} (t^{- 2} ln (t))